You only have to remember it:
$\displaystyle a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$

Sep 1st 2009, 09:38 AM

RobLikesBrunch

Quote:

Originally Posted by ynj

You only have to remember it:
$\displaystyle a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$

Simply memorizing mathematics without understanding is absurd and idiotic. It is much better to not memorize a certain formula and derive it each time...even if it takes you longer.

Sep 1st 2009, 11:27 AM

Taluivren

Quote:

Originally Posted by RobLikesBrunch

Simply memorizing mathematics without understanding is absurd and idiotic.

(Giggle) agree with you, but some things are good to know. I don't remember this formula, but what initially helped me to factor the expression was the formula $\displaystyle x^3+y^3 = (x+y)(x^2+y^2 -xy)$.
So here is the other way how to factorize, and pacman you'll find another ways if you try.

I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down. Thanks for the efforts to help. (Clapping)(Clapping)(Clapping)(Clapping)(Clapping)

Sep 2nd 2009, 12:17 AM

Taluivren

Quote:

Originally Posted by pacman

I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down.

for me, it is about messing around with symbols using identities i already know and recognizing patterns and symmetries. Maybe we should start to learn from computers, i suppose they are better in this than humans. Can anybody explain how computers do it? (i mean how they factorize (Wink))

Sep 2nd 2009, 02:03 AM

pacman

i found this from the web, now i wonder why it is difficult to factor

The identity would probably be known to Lagrange from his extensive study of algebraic equations If w is a primitive cubic root of unity then a^3 + b^3 + c^3 - 3abc is the constant term of the polynomial satisfied by a +bw +cww
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a +bw +cww)(a +bww +cw).

This and other similar identities occur when symmetrical functions of the roots of polynomial
equations are calculated I know Newton studied symmetric functions of roots. He may have been aware of this identity.

The (a +bw +cww) is a special form of Eisenstein cubic integrers and a^3 + b^3 + c^3 - 3abc ist its norm. thus, a^3 + b^3 + c^3 - 3abc can also be referrred to a a termary cubic form.